Circle geometry

Use the theorem about the perpendicular lines from the centre of a circle.

We need to determine the length of $AT$. We know that $AB$ has a length of $40$. To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

The segment $OT$ is perpendicular to chord $AB$. Therefore $OT$ bisects $AB$. And that means $AT$ is half as long as $AB$. We can use that fact to find the length of $AT$.

$AT$ has a length of $20$.

Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.

We need to find the radius of the circle (the length of $OA$). Let's start by labelling the information we know about the figure, and the radius that we want to find. In $\Delta OSA$, side $SA$ must be half as long as $AC$. This is for the same reason we used in Question 1. So $SA = 24$, as labelled below. We are using $d$ for the length of $OT$.

There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In $\Delta OSA$:

We can write a similar equation for $\Delta OTA$.

Now we have two equations that are both equal to $r^{2}$. So they must be equal to each other, and we can write:

Now solve the equation for $d$ (which is the length of $OT$).

The result we want is a distance, so we can throw away the negative answer.

We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

Again, we throw away the negative answer because we are calculating a distance. The length of radius $OA$ is $25$.

Use the theorem about the perpendicular lines from the centre of a circle.

We need to determine the length of $AT$. We know that $AB$ has a length of $40$. To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

The segment $OT$ is perpendicular to chord $AB$. Therefore $OT$ bisects $AB$. And that means $AT$ is half as long as $AB$. We can use that fact to find the length of $AT$.

$AT$ has a length of $20$.

Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.

We need to find the radius of the circle (the length of $OA$). Let's start by labelling the information we know about the figure, and the radius that we want to find. In $\Delta OSA$, side $SA$ must be half as long as $AC$. This is for the same reason we used in Question 1. So $SA = 24$, as labelled below. We are using $d$ for the length of $OT$.

There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In $\Delta OSA$:

We can write a similar equation for $\Delta OTA$.

Now we have two equations that are both equal to $r^{2}$. So they must be equal to each other, and we can write:

Now solve the equation for $d$ (which is the length of $OT$).

The result we want is a distance, so we can throw away the negative answer.

We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

Again, we throw away the negative answer because we are calculating a distance. The length of radius $OA$ is $25$.

Use the theorem about the perpendicular lines from the centre of a circle.

We need to determine the length of $AX$. We know that $AC$ has a length of $154$. To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

The segment $OX$ is perpendicular to chord $AC$. Therefore $OX$ bisects $AC$. And that means $AX$ is half as long as $AC$. We can use that fact to find the length of $AX$.

$AX$ has a length of $77$.

Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.

We need to find the radius of the circle (the length of $OA$). Let's start by labelling the information we know about the figure, and the radius that we want to find. In $\Delta OYA$, side $YA$ must be half as long as $AB$. This is for the same reason we used in Question 1. So $YA = 68$, as labelled below. We are using $d$ for the length of $OY$.

There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In $\Delta OXA$:

We can write a similar equation for $\Delta OYA$.

Now we have two equations that are both equal to $r^{2}$. So they must be equal to each other, and we can write:

Now solve the equation for $d$ (which is the length of $OY$).

The result we want is a distance, so we can throw away the negative answer.

We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

Again, we throw away the negative answer because we are calculating a distance. The length of radius $OA$ is $85$.

We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'

One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

In this figure, $\overline{AP}$ is the chord, and it is bisected at point $G$. In other words, $G$ is the midpoint of $\overline{AP}$. We know the chord is bisected because it is perpendicular to segment $\overline{OG}$.

On the diagram below the segments $\overline{AG}$ and $\overline{GP}$ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

Based on this, we can get the answers: $\overline{AP}$ is "twice as long as $\overline{GP}$" because "$\perp$ line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).

If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!

The length that we want is bigger than the length given. Therefore, we must multiply by two to get the length of the chord.

The length of $\overline{AP}$ is $\text{22}$.

We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'

One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

In this figure, $\overline{KH}$ is the chord, and it is bisected at point $C$. In other words, $C$ is the midpoint of $\overline{KH}$. We know the chord is bisected because it is perpendicular to segment $\overline{OC}$.

On the diagram below the segments $\overline{KC}$ and $\overline{CH}$ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

Based on this, we can get the answers: $\overline{KH}$ is "twice as long as $\overline{KC}$" because "$\perp$ line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).

If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!

The length that we want is bigger than the length given. Therefore, we must multiply by two to get the length of the chord.

The length of $\overline{KH}$ is $\text{36}$.

We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'

One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

In this figure, $\overline{DF}$ is the chord, and it is bisected at point $Q$. In other words, $Q$ is the midpoint of $\overline{DF}$. We know the chord is bisected because it is perpendicular to segment $\overline{OQ}$.

On the diagram below the segments $\overline{DQ}$ and $\overline{QF}$ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

Based on this, we can get the answers: $\overline{QF}$ is "half the length of $\overline{DF}$" because "$\perp$ line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).

If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!

The length that we want is smaller than the length given. Therefore, we must divide by two to get the length of the smaller segment (this is the same as multiplying by a half).

The length of $\overline{QF}$ is $\text{4}$.

To name a quadrilateral, you must name all four of its vertices: pick any vertex of the quadrilateral and then go around the perimeter of the quadrilateral and add each vertex as you come to it.

The figure shows the four points which lie on the circle. They are the vertices of the quadrilateral (they are circled in red below). Naming the cyclic quadrilateral means that we must name all four of those vertices as we go around the perimeter of the quadrilateral. (We can ignore point $P$ because it is not on the circle.)

The answer can be any list of the four points on the circumference of the circle, as long as they follow the order of the points around the circle. Any of these answers is correct:

Remember that we know a theorem about the opposite angles of a cyclic quadrilateral: they are supplementary.

We must use this fact about cyclic quadrilaterals: their opposite angles are supplementary (have a sum of $\text{180}$ degrees). We can use that fact to answer the question.

The quadrilateral has two pairs of opposite angles: $\hat{Q}$ and $\hat{S}$ are opposite each other; $\hat{R}$ and $\hat{T}$ are opposite from each other.

We know that $\hat{R}$ and $\hat{T}$ have a sum of $180°$:

It is the same story over again for angles $\hat{S}$ and $\hat{Q} \longrightarrow \hat{Q} = 180 - 109° = 71°$.

The answers are:

1. The answer for $S\hat{T}Q$ is $108°$.
2. The answer for $T\hat{Q}R$ is $71°$.

Theorems state a connection between two things, and the converse states the same connection 'backwards.' Does the work you did above use the following connection straight through, or does it go backwards: 'Opposite angles in a cyclic quadrilateral are supplementary.'

The theorem about the angles in cyclic quadrilaterals states the following connection:

• Theorem: If a quadrilateral is cyclic $\longrightarrow$ the opposite angles must be supplementary.

The converse states the same connection but in 'reverse:'

• Converse: If the opposite angles of a quadrilateral are supplementary $\longrightarrow$ the quadrilateral must be cyclic.

Solving the questions above uses the theorem: because we know the quadrilateral is cyclic, then we also know that the angles are supplementary.

We solve Question 2 using a theorem.

To name a quadrilateral, you must name all four of its vertices: pick any vertex of the quadrilateral and then go around the perimeter of the quadrilateral and add each vertex as you come to it.

The figure shows the four points which lie on the circle. They are the vertices of the quadrilateral (they are circled in red below). Naming the cyclic quadrilateral means that we must name all four of those vertices as we go around the perimeter of the quadrilateral. (We can ignore point $C$ because it is not on the circle.)

The answer can be any list of the four points on the circumference of the circle, as long as they follow the order of the points around the circle. Any of these answers is correct:

Remember that we know a theorem about the opposite angles of a cyclic quadrilateral: they are supplementary.

We must use this fact about cyclic quadrilaterals: their opposite angles are supplementary (have a sum of $\text{180}$ degrees). We can use that fact to answer the question.

The quadrilateral has two pairs of opposite angles: $\hat{A}$ and $\hat{D}$ are opposite each other; $\hat{B}$ and $\hat{E}$ are opposite from each other.

We know that $\hat{E}$ and $\hat{B}$ have a sum of $180°$:

It is the same story over again for angles $\hat{A}$ and $\hat{D} \longrightarrow \hat{D} = 180 - 61° = 119°$.

The answers are:

1. The answer for $A\hat{B}D$ is $80°$.
2. The answer for $B\hat{D}E$ is $119°$.

Theorems state a connection between two things, and the converse states the same connection 'backwards.' Does the work you did above use the following connection straight through, or does it go backwards: 'Opposite angles in a cyclic quadrilateral are supplementary.'

The theorem about the angles in cyclic quadrilaterals states the following connection:

• Theorem: If a quadrilateral is cyclic $\longrightarrow$ the opposite angles must be supplementary.

The converse states the same connection but in 'reverse:'

• Converse: If the opposite angles of a quadrilateral are supplementary $\longrightarrow$ the quadrilateral must be cyclic.

Solving the questions above uses the theorem: because we know the quadrilateral is cyclic, then we also know that the angles are supplementary.

We solve Question 2 using a theorem.

To name a quadrilateral, you must name all four of its vertices: pick any vertex of the quadrilateral and then go around the perimeter of the quadrilateral and add each vertex as you come to it.

The figure shows the four points which lie on the circle. They are the vertices of the quadrilateral (they are circled in red below). Naming the cyclic quadrilateral means that we must name all four of those vertices as we go around the perimeter of the quadrilateral. (We can ignore point $I$ because it is not on the circle.)

The answer can be any list of the four points on the circumference of the circle, as long as they follow the order of the points around the circle. Any of these answers is correct:

Remember that we know a theorem about the opposite angles of a cyclic quadrilateral: they are supplementary.

We must use this fact about cyclic quadrilaterals: their opposite angles are supplementary (have a sum of $\text{180}$ degrees). We can use that fact to answer the question.

The quadrilateral has two pairs of opposite angles: $\hat{G}$ and $\hat{J}$ are opposite each other; $\hat{H}$ and $\hat{K}$ are opposite from each other.

We know that $\hat{J}$ and $\hat{G}$ have a sum of $180°$:

It is the same story over again for angles $\hat{K}$ and $\hat{H} \longrightarrow \hat{H} = 180 - 96° = 84°$.

The answers are:

1. The answer for $K\hat{G}H$ is $49°$.
2. The answer for $G\hat{H}J$ is $84°$.

Theorems state a connection between two things, and the converse states the same connection 'backwards.' Does the work you did above use the following connection straight through, or does it go backwards: 'Opposite angles in a cyclic quadrilateral are supplementary.'

The theorem about the angles in cyclic quadrilaterals states the following connection:

• Theorem: If a quadrilateral is cyclic $\longrightarrow$ the opposite angles must be supplementary.

The converse states the same connection but in 'reverse:'

• Converse: If the opposite angles of a quadrilateral are supplementary $\longrightarrow$ the quadrilateral must be cyclic.

Solving the questions above uses the theorem: because we know the quadrilateral is cyclic, then we also know that the angles are supplementary.

We solve Question 2 using a theorem.

You may find it helpful to revise circle geometry in the Everything Maths textbook.

Use the statement from Question 1 to start the proof.

The table below shows the completed proof, that opposite angles of a cyclic quadrilateral are supplementary (add up to $180°$).

You may find it helpful to revise circle geometry in the Everything Maths textbook.

Use the statement from Question 1 to start the proof.

The table below shows the completed proof, that opposite angles of a cyclic quadrilateral are supplementary (add up to $180°$).

You may find it helpful to revise circle geometry in the Everything Maths textbook.

Use the statement from Question 1 to start the proof.

The table below shows the completed proof, that opposite angles of a cyclic quadrilateral are supplementary (add up to $180°$).

'Cyclic' means 'on a circle' or 'following a circular path.' A quadrilateral is a shape with four sides. Together, the words 'cyclic quadrilateral' mean a shape with four sides and with vertices sitting on a circle.

Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle (concyclic). We need to focus on the points which are on the circumference of the circle: for example, we can ignore point $R$ because it is not on the circle.

The figure shows 2 points on the circle, which are circled in red. (Point $Q$ is inside of the circle, not on the circumference!)

There is no cyclic shape in this figure.

The correct answer from the list is: there is no cyclic quadrilateral.

You need to know the following about cyclic quadrilaterals: (a) the vertices of the quadrilateral must sit on the edge (circumference) of a circle, and (b) opposite angles of a cyclic quadrilateral are supplementary (they have a sum of 180 degrees).

The correct answer is opposite angles have a sum of 180 degrees.

'Cyclic' means 'on a circle' or 'following a circular path.' A quadrilateral is a shape with four sides. Together, the words 'cyclic quadrilateral' mean a shape with four sides and with vertices sitting on a circle.

Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle (concyclic). We need to focus on the points which are on the circumference of the circle: for example, we can ignore point $G$ because it is not on the circle.

The figure shows 4 points on the circle, which are circled in red.

Because the shape has four sides and the vertices are all on the edge of the circle, the figure $HIJK$ is a cyclic quadrilateral. The cyclic quadrilateral is shown shaded in light blue.

The correct answer from the list is: HIJK.

The quadrilateral has two pairs of opposite angles: $\hat{H}$ and $\hat{J}$ are opposite each other; $\hat{I}$ and $\hat{K}$ are opposite from each other.

The correct answer is J.

'Cyclic' means 'on a circle' or 'following a circular path.' A quadrilateral is a shape with four sides. Together, the words 'cyclic quadrilateral' mean a shape with four sides and with vertices sitting on a circle.

Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle (concyclic). We need to focus on the points which are on the circumference of the circle: for example, we can ignore point $B$ because it is not on the circle.

The figure shows 3 points on the circle, which are circled in red.

There is a cyclic triangle in this figure, but not a cyclic quadrilateral.

The correct answer from the list is: there is no cyclic quadrilateral.

You need to know the following about cyclic quadrilaterals: (a) the vertices of the quadrilateral must sit on the edge (circumference) of a circle, and (b) opposite angles of a cyclic quadrilateral are supplementary (they have a sum of 180 degrees).

The correct answer is opposite angles have a sum of 180 degrees.

A good start for any geometry question is to add all information you know about the shapes in the figure: are there any things you can label in this figure? Indeed, the question tells you the radius of the circle is $2 c + 35$, as well as the length of the line segment $\overline{HI} = c + 55$. Put this information on the diagram - when you label the diagram it helps to organise the information.

There are also some congruent (equal) line segments in the figure: you should mark them to show that the segments are equal in length. One pair of equal segments are the radii (see below).

Now think about what the question is asking you to find: you must determine the length of $\overline{GI}$. What comparisons can you make to determine the length of that segment? There is a theorem about tangent lines with circles which might be helpful: if two tangent lines to a circle come from the same point, then they have equal lengths.

There are two tangent lines in this figure, and they come from the same place (point $I$). Therefore the segments are congruent: $\overline{GI} \equiv \overline{HI}$ (the symbol $\equiv$ means 'congruent'). Therefore, the answer is $\overline{GI} = c + 55$.

In maths, and in Euclidean geometry in particular, it is important to know the reason that allows you to find some piece of information. In this question, if the reason is one of the theorems you need to know which one it is!

As described above, the correct answer is: Two tangents drawn from the same point outside a circle.

A good start for any geometry question is to add all information you know about the shapes in the figure: are there any things you can label in this figure? Indeed, the question tells you the radius of the circle is $\frac{3 y}{4} - 75$, as well as the length of the line segment $\overline{VX} = \frac{y}{2} - 20$. Put this information on the diagram - when you label the diagram it helps to organise the information.

Now think about what the question is asking you to find: the question asks for the length of the diameter. The diameter is always twice as long as the radius: $d = 2r$.

To determine the answer, you just need to take the radius, which is given, and double it: $d = 2 \left( \frac{3 y}{4} - 75 \right) = \frac{3 y}{2} - 150$.

In maths, and in Euclidean geometry in particular, it is important to know the reason that allows you to find some piece of information. In this question, if the reason is one of the theorems you need to know which one it is!

However, this question does not use any special theorem: you just use the fact that the diameter is twice as long as the radius. The correct answer is: 'None of the above.'

A good start for any geometry question is to add all information you know about the shapes in the figure: are there any things you can label in this figure? Indeed, the question tells you the radius of the circle is $\frac{4 y}{3} + 35$, as well as the length of the line segment $\overline{VX} = \frac{3 y}{2} + 80$. Put this information on the diagram - when you label the diagram it helps to organise the information.

There are also some congruent (equal) line segments in the figure: you should mark them to show that the segments are equal in length. One pair of equal segments are the radii (see below).

Now think about what the question is asking you to find: you must determine the length of $\overline{WX}$. What comparisons can you make to determine the length of that segment? There is a theorem about tangent lines with circles which might be helpful: if two tangent lines to a circle come from the same point, then they have equal lengths.

There are two tangent lines in this figure, and they come from the same place (point $X$). Therefore the segments are congruent: $\overline{WX} \equiv \overline{VX}$ (the symbol $\equiv$ means 'congruent'). Therefore, the answer is $\overline{WX} = \frac{3 y}{2} + 80$.

In maths, and in Euclidean geometry in particular, it is important to know the reason that allows you to find some piece of information. In this question, if the reason is one of the theorems you need to know which one it is!

As described above, the correct answer is: Two tangents drawn from the same point outside a circle.

Use the parallel lines in the figure. Notice that $PT$ is a transversal.

The figure includes the parallel lines $RT$ and $PQ$. The line $PT$ is a transversal crossing these parallel lines. So we can use what we know about transversal geometry. This might include corresponding angles, alternate interior angles, and so on.

In this case, $\hat{P}_{2}$ and $\hat{T}_{1}$ are alternate interior angles. And alternate interior angles on parallel lines are equal:

$\hat{T}_{1}$ is given in the question as $70°$.

$\hat{P}_{2}$ must be equal to $70°$ because alternate interior angles are equal. The correct answer is Option A.

You can find $\hat{ Q }_{1}$ using the fact that triangle $PQT$ is isosceles.

From Question 1 we know that $\hat{ P }_ { 2 } = 70°$. We also know from the question that triangle $PQT$ is isosceles with $\hat{P}_{2}$ equal to $\hat{T}_{2}$. So that means $\hat{T}_{2}$ must also be $70°$.

Now we know two of the angles in triangle $PQT$ so we can calculate the size of $\hat{ Q }_{1}$:

To find both angles in triangle $PQT$, we used the sum of angles in a triangle.

At Point $T$ there are three angles, and they make a straight line. So they must have a sum of $180°$.

We know from the diagram that $\hat{ T }_{ 1 } = 70°$. And from the work above we know that $\hat{ T }_{ 2 }$ is also equal to $70°$.

The key reason for this information is angles on a straight line.

Use the parallel lines in the figure. Notice that $QR$ is a transversal.

The figure includes the parallel lines $PQ$ and $RT$. The line $QR$ is a transversal crossing these parallel lines. So we can use what we know about transversal geometry. This might include corresponding angles, alternate interior angles, and so on.

In this case, $\hat{R}_{1}$ and $\hat{Q}_{1}$ are alternate interior angles. And alternate interior angles on parallel lines are equal:

$\hat{Q}_{1}$ is given in the question as $69°$.

$\hat{R}_{1}$ must be equal to $69°$ because alternate interior angles are equal. The correct answer is Option C.

You can find $\hat{ T }_{1}$ using the fact that triangle $QRT$ is isosceles.

From Question 1 we know that $\hat{ R }_ { 1 } = 69°$. We also know from the question that triangle $QRT$ is isosceles with $\hat{R}_{1}$ equal to $\hat{Q}_{2}$. So that means $\hat{Q}_{2}$ must also be $69°$.

Now we know two of the angles in triangle $QRT$ so we can calculate the size of $\hat{ T }_{1}$:

To find both angles in triangle $QRT$, we used the sum of angles in a triangle.

The fact that $RS$ is a tangent line at Point $R$ means we can consider theorems about tangent lines. According to the tangent chord theorem, the angle between a tangent and a chord (in this case, the chord is $QR$) is equal to an angle on the circle's circumference subtended by that chord. In this diagram, that means $\hat{ R }_{ 2 }$ is equal to $\hat{ T }_{1}$:

We know from our work above that $\hat{ T }_{1} = 42°$. So:

The key reason for this information is the tangent chord theorem.

Use the parallel lines in the figure. Notice that $AE$ is a transversal.

The figure includes the parallel lines $CE$ and $AB$. The line $AE$ is a transversal crossing these parallel lines. So we can use what we know about transversal geometry. This might include corresponding angles, alternate interior angles, and so on.

In this case, $\hat{A}_{1}$ and $\hat{E}_{1}$ are alternate interior angles. And alternate interior angles on parallel lines are equal:

$\hat{E}_{1}$ is given in the question as $72°$.

$\hat{A}_{1}$ must be equal to $72°$ because alternate interior angles are equal. The correct answer is Option B.

You can find $\hat{ B }_{1}$ using the fact that triangle $ABE$ is isosceles.

From Question 1 we know that $\hat{ A }_ { 1 } = 72°$. We also know from the question that triangle $ABE$ is isosceles with $\hat{A}_{1}$ equal to $\hat{E}_{2}$. So that means $\hat{E}_{2}$ must also be $72°$.

Now we know two of the angles in triangle $ABE$ so we can calculate the size of $\hat{ B }_{1}$:

To find both angles in triangle $ABE$, we used the sum of angles in a triangle.

At Point $E$ there are three angles, and they make a straight line. So they must have a sum of $180°$.

We know from the diagram that $\hat{ E }_{ 1 } = 72°$. And from the work above we know that $\hat{ E }_{ 2 }$ is also equal to $72°$.

The key reason for this information is angles on a straight line.

Sometimes geometry figures can be very busy! It is a good idea to look at some parts of the pictures separately from the rest. You may even want to use your fingers to cover up some lines: this will help to simplify the picture. Move your hand around different places and see if it helps.

The word 'congruent' means identical: same size, shape, etc. To find two congruent line segments, we need to find two line segments with the same length.

One of the circle theorems says that if there are two tangent lines to a circle from the same point, then they must be the same length. Are there two tangent lines like that in this figure?

To test if there are two equal tangent lines, start at a point outside the circle: then look to see if there are two lines from that point stretching toward the circle... and if they are both tangent to the circle. If there are not two lines, or if the are not both tangent lines, then we cannot know that there are equal segments in the diagram.

You can see there are two different points outside of the circle: $I$ and $L$. Check each of them to see if there are two lines reaching the circle, and if those lines are tangent to the circle.

In this case, the diagram does include a pair of tangent lines: they pass from point $I$ and are tangent to the circle at points $K$ and $J$. You can see these segments shown in red in the diagram above. They are also marked as congruent.

The correct answer is: $IJ; IK$.

Remember that a tangent line is always perpendicular to the radius reaching the tangent point... In this figure, is there a radius in the circle which touches a tangent point? If so, then there is a right angle there!

One thing you must remember about tangent lines for circles is this fact: the tangent line will be perpendicular to the radius touching the tangent point.

Any radius must come from the centre of the circle. Are there any lines starting at point $O$ in this figure? The red line segment shown starts at the centre of the circle, but at the spot where it touches the circle, there is no tangent line.

Based on what you can see in the diagram, the answer is none.

Sometimes geometry figures can be very busy! It is a good idea to look at some parts of the pictures separately from the rest. You may even want to use your fingers to cover up some lines: this will help to simplify the picture. Move your hand around different places and see if it helps.

The word 'congruent' means identical: same size, shape, etc. To find two congruent line segments, we need to find two line segments with the same length.

One of the circle theorems says that if there are two tangent lines to a circle from the same point, then they must be the same length. Are there two tangent lines like that in this figure?

To test if there are two equal tangent lines, start at a point outside the circle: then look to see if there are two lines from that point stretching toward the circle... and if they are both tangent to the circle. If there are not two lines, or if the are not both tangent lines, then we cannot know that there are equal segments in the diagram.

You can see there are two different points outside of the circle: $R$ and $W$. Check each of them to see if there are two lines reaching the circle, and if those lines are tangent to the circle.

In this case, the diagram does include a pair of tangent lines: they pass from point $R$ and are tangent to the circle at points $P$ and $Q$. You can see these segments shown in red in the diagram above. They are also marked as congruent.

The correct answer is: $QR; PR$.

Remember that a tangent line is always perpendicular to the radius reaching the tangent point... In this figure, is there a radius in the circle which touches a tangent point? If so, then there is a right angle there!

One thing you must remember about tangent lines for circles is this fact: the tangent line will be perpendicular to the radius touching the tangent point.

Any radius must come from the centre of the circle. Are there any lines starting at point $O$ in this figure? The red line segment shown starts at the centre of the circle, but at the spot where it touches the circle, there is no tangent line.

Based on what you can see in the diagram, the answer is none.

Sometimes geometry figures can be very busy! It is a good idea to look at some parts of the pictures separately from the rest. You may even want to use your fingers to cover up some lines: this will help to simplify the picture. Move your hand around different places and see if it helps.

The word 'congruent' means identical: same size, shape, etc. To find two congruent line segments, we need to find two line segments with the same length.

One of the circle theorems says that if there are two tangent lines to a circle from the same point, then they must be the same length. Are there two tangent lines like that in this figure?

To test if there are two equal tangent lines, start at a point outside the circle: then look to see if there are two lines from that point stretching toward the circle... and if they are both tangent to the circle. If there are not two lines, or if the are not both tangent lines, then we cannot know that there are equal segments in the diagram.

You can see there are two different points outside of the circle: $B$ and $E$. Check each of them to see if there are two lines reaching the circle, and if those lines are tangent to the circle.

In this case, the diagram does include a pair of tangent lines: they pass from point $B$ and are tangent to the circle at points $D$ and $C$. You can see these segments shown in red in the diagram above. They are also marked as congruent.

The correct answer is: $BC; BD$.

Remember that a tangent line is always perpendicular to the radius reaching the tangent point... In this figure, is there a radius in the circle which touches a tangent point? If so, then there is a right angle there!

One thing you must remember about tangent lines for circles is this fact: the tangent line will be perpendicular to the radius touching the tangent point.

Any radius must come from the centre of the circle. Are there any lines starting at point $O$ in this figure? Yes, there is a radius which goes to point $C$; see the red radius below.

Based on what you can see in the diagram, the answer is $B\hat{C}O$; this can also be written as $O\hat{C}B$.